Need help researching if these series are divergent or convergent

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[woeful finch]

I've used D'alambert and Raabe's criterions, but I always get 1

[past vale]

Note:
(2n - 1)!! = (2n - 1)!/(2^(n - 1) (n - 1)!)
(2n)!! = 2^n n!
And also recall Stirling's formula.

[woeful finch]

(2n-1)!!=(2n)!/2^n(n)!

yeah?

[past vale]

That is (2n + 1)!!.

In your case the numerator should be (2n + 1)!, by the way.

[woeful finch]

really?
(2n-1)!!=(2n-1)(2n-3)...

multiplying numerator and denominator by (2n)!!=(2n)(2n-2)(2n-4)
(2n)(2n-1)(2n-2)(2n-3).../(2n)(2n-2)(2n-4)...
(2n)!/(2^n)n! checks out

[past vale]

Ah, I see.
(2n - 1)!! = (2n - 1)!/(2^(n - 1) (n - 1)!) = (2n)!/(2^n n!)
Both forms work, my bad.

[woeful finch]

all good

problem is

we haven't studied Stirling's formula

but at this point anything is a solution to me since I've tried many times

I need like the formula and an example of how it can be done

sicne I have no idea

[woeful finch]

so for a_n I have

how do I use Stirling's formula?

this is Stirling, right?

[past vale]

Well, you only need n! ~ √(2πn)(n/e)^n for n -> ∞.

[woeful finch]

right, so for (2n)! it should become:

right?

[past vale]

Yes.

[woeful finch]

but what about 2^(2n)?

I don't work with it?

[past vale]

That remains for now, but after you use Stirling's formula for n!^2 in the denominator, you'll see what happens.

[woeful finch]

this?

[past vale]

Yeah.

Oh, by the way, don't forget that everything should also be squared.

[woeful finch]

yeah I'm just working with the expression inside the brackets

[past vale]

Ah, ok.

[woeful finch]

this is true?

[past vale]

Of course.

[woeful finch]

so then I get this

[past vale]

√(4πn) = 2√(πn), not 2√(2πn). Other than that, yes. It should become 1/√(πn).

[woeful finch]

ohh

[fast snow]

does something like say root tests not work

totally

[woeful finch]

I just got a word from my lecturer

he said to use Gaus' criterion

but we haven't studied it yet, so I'd need help

[past vale]

Oh, Gauss? Interesting.

[fast snow]

the ratio of the terms here is

[past vale]

Ah, it works like this.
Let a(n) be the term of the series and suppose we have the following (λ > 1):
a(n)/a(n + 1) = A + B/n + C/n^λ
Then:

1. If A > 1, the series converges, if A < 1 - diverges.
2. A = 1: if B > 1, the series converges, if B ≤ 1 - diverges.

Actually, maybe Raabe is enough? Let me try.

[woeful finch]

nah I tried Raabe 5 times already

[fast snow]

a(n)/a(n+1)=((2n+2)/(2n+1))^2
(1 + 1/(2n+1))^2
= 1 + 2/(2n+1) + 1/(2n+1)^2

hmm it isn't in "n"

but 2n+1

[past vale]

a(n) = ((2n - 1)!!/(2n)!!)^2
a(n + 1) = ((2n + 1)!!/(2n + 2)!!)^2
a(n)/a(n + 1) = ((2n + 2)/(2n + 1))^2 = 1 + 1/(n + 1/2) + (1/4)/(n + 1/2)^2
Yeah, Raabe won't work. But we do get divergence by Gauss: A = 1, B ≤ 1.

Interesting! I haven't used Gauss in ages.

[fast snow]

gaus should

[past vale]

Hm. Just for fun, let me try Bertrand, too.

[fast snow]

yeah no it ain't gonna work

[woeful finch]

yeah I need hel pwith Gauss 😭

it's for an assignment

I gotta write it up

[fast snow]

a(n+1)/a(n) = ((2n+1)/(2n+1))^2

does there exist some b > 1

a(n+1)/a(n) <= 1-b/n

[past vale]

a(n)/a(n + 1) - 1 = 1/(n + 1/2) + (1/4)/(n + 1/2)^2
n(a(n)/a(n + 1) - 1) = n/(n + 1/2) + (1/4)n/(n + 1/2)^2 = 1 - (1/4)/(n + 1/2) - (1/8)/(n + 1/2)^2
n(a(n)/a(n + 1) - 1) - 1 = -(1/4)/(n + 1/2) - (1/8)/(n + 1/2)^2
ln(n)(n(a(n)/a(n + 1) - 1) - 1) = -ln(n)((1/4)/(n + 1/2) + (1/8)/(n + 1/2)^2)
This goes to 0 when n -> ∞, so Bertrand's test also works.

[fast snow]

why is gaus test 1 + a/n + O(1/n^p)

could've been 1 + O(1/n) + O(1/n^p)

[past vale]

The costant of 1/n matters.

I wrote it above.

[fast snow]

right

[past vale]

I think there was a generalized version of Gauss, too.

[woeful finch]

they wrote it like this

[past vale]

Oh, they are generalizing it, too.

Well, yeah. Then you look at the first couple of terms and you can see when it converges by Gauss and when it diverges.

[woeful finch]

damn nice

[woeful finch]

uhh so

help me understand

this is true right?

so alpha and beta are just some random numbers

but if alpha is lesser or equals to one, series diverge

ok so far so good

but how do I connect:

[past vale]

Note that 2n + 1 = 2(n + 1/2).

And n + 1/2 is pretty much the same as n for large n.

[woeful finch]

right

so

[past vale]

Yes. In other, words, a(n)/a(n + 1) = 1 + 1/(n + 1/2) + (1/4)/(n + 1/2)^2. So, now look at what I wrote about Gauss's test and see what you can say here.

[woeful finch]

divergent!

alpha = 1 , beta <= 1

[past vale]

Yes, very good! And what about your general case with the power p?

[woeful finch]

idk 😭

p = 2

[past vale]

Well, let's see. We have:
a(n)/a(n + 1) = (1 + 1/(2n + 1))^p
Try expanding this using the usual binomial identity.

You don't need all terms, just stop at the quadratic one.

[woeful finch]

idk really I don't want to be bothered

I've spent over 2 full days

I'm done

[past vale]

Fine, let me show. It's very easy.

[woeful finch]

alr

[past vale]

We have:
(1 + 1/(2n + 1))^p = 1 + p/(2n + 1) + O(1/(2n + 1)^2) = 1 + (p/2)/(n + 1/2) + O(1/(2n + 1)^2)
By Gauss, this converges when p/2 > 1, so when p > 2. So, the series with a(n) = ((2n - 1)!!/(2n)!!)^p converges for p > 2 and diverges for p ≤ 2.

Our particular case was p = 2, which diverges.

[woeful finch]

yeah p/2 = 2/2 <=1

[past vale]

Hm... Let me just quickly search for generalized Gauss test, so I don't forget.

Yeah, here it is! Translation (row by row):
Suppose the series with term a(n) > 0 satisfies...
Then if...
...then the series converges. Otherwise, it diverges.
@fast snow , you might also be interested in this.

And I think you can see how this can be generalized as far as you want.

I wonder if there is a series that this won't work on no matter how many terms we take.

[woeful finch]

yeah that's good

I gotta go now tho

thanks for the help!

[late pike]

Unable to parse the channel name

[past vale]

You're welcome!